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# Answer to Question #85285 in Algorithms for Mlk

Question #85285
Prove that n + log2n = O(n) by showing that there exists a constant c &gt; 0 such that n + log2n &le; cn. (note that log2n means (log n)2.)
1
2019-02-22T05:38:07-0500
@$(\log^2⁡n-n)'=2 \log ⁡n \times 1/n-1=2 ((\log ⁡n-n))/n\le0,\quad n\in\mathbb{N}@$@$(\log ⁡n-n)'=1/n-1\le0@$@$\log^2 n \le n @$@$n + \log^2 n \le n + n = 2n,\quad n\in \mathbb{N}@$

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